題目鏈接： http://poj.org/problem?id=2777
Description
There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:?
1. "C A B C" Color the board from segment A to segment B with color C.?
2. "P A B" Output the number of different colors painted between segment A and segment B (including).?
In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.?
Input
Output
Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
Source
題意：
給一個固定長度為L的畫板
有兩個操作：
C A B C：區間A--B內涂上顏色C。
P A B：查詢區間AB內顏色種類數。
PS:此題和 HDU：5023 是類似的！
附題解： http://blog.csdn.net/u012860063/article/details/39434665
代碼例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define LL int
const int maxn = 110017;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
//把當前結點的信息更新到父結點
sum[rt] = sum[rt<<1] | sum[rt<<1|1];//總共的顏色
}
void PushDown(int rt,int m)
{
if(add[rt])
{
add[rt<<1] = add[rt];
add[rt<<1|1] = add[rt];
sum[rt<<1] = add[rt];
sum[rt<<1|1] = add[rt];
add[rt] = 0;//將標記向兒子節點移動后，父節點的延遲標記去掉
//傳遞后，當前節點標記域清空
}
}
void build(int l,int r,int rt)
{
add[rt] = 0;//初始化為全部結點未被標記
if (l == r)
{
sum[rt] = 1;//初始顏色為1
return ;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if (L <= l && r <= R)
{
add[rt] =1<<(c-1);//位運算左移表示有某種顏色
sum[rt] =1<<(c-1);
return ;
}
PushDown(rt , r - l + 1);//----延遲標志域向下傳遞
int mid = (l + r) >> 1;
if (L <= mid)
update(L , R , c , lson);//更新左兒子
if (mid < R)
update(L , R , c , rson);//更新右兒子
PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
if (L <= l && r <= R)
{
return sum[rt];
}
//要取rt子節點的值時，也要先把rt的延遲標記向下移動
PushDown(rt , r - l + 1);
int mid = (l + r) >> 1;
LL ret = 0;
if (L <= mid)
ret |= query(L , R , lson);
if (mid < R)
ret |= query(L , R , rson);
return ret;
}
int main()
{
int L, T, O;
int a, b, c;
while(~scanf("%d%d%d",&L,&T,&O))
{
build(1, L, 1);//建樹
while(O--)//Q為詢問次數
{
char op[2];
scanf("%s",op);
if(op[0] == 'P')
{
scanf("%d%d",&a,&b);
if(a > b)
{
int t = a;
a = b;
b = t;
}
LL tt=query(a, b, 1, L, 1);
int ans = 0;
while(tt)
{
if(tt&1)
{
ans++;
}
tt>>=1;
}
printf("%d\n",ans);
}
else
{
scanf("%d%d%d",&a,&b,&c);
if(a > b)
{
int t = a;
a = b;
b = t;
}
update(a, b, c, 1, L, 1);
}
}
}
return 0;
}
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